How do you graph, find any intercepts, domain and range of #f(x)=(1/4)^(x+3)-2#?

1 Answer
Apr 10, 2017

Doman is #(-oo,oo)#, range is #(-2,oo)#, #y#-intercept is #-1 63/64# and #x#-intercept is #-7/2#.

Explanation:

#f(x)=(1/4)^(x+3)-2#

#=(4^(-1))^(x+3)-2#

#=(2^2)^(-x-3)-2#

#=2^(-2x-6)-2#

As any power of #2# is possible, there are no limitations on value of #x# and doman is #(-oo,oo)#

However, as #x->oo#, as #2^(-oo)->0#, value of #f(x)# cannot go below #-2# and range is #(-2,oo)#

Further when #x=0#, #f(x)=2^(-6)-2=1/64-2=-1 63/64# and hence #y#-intercept is #-1 63/64# and as #f(x)=0# is at #2^(-2x-6)=2# or #-2x-6=1# i.e. #x=-7/2#, #x#-intercept is #-7/2#.

The graph appears as follows.

graph{(1/4)^(x+3)-2 [-10, 10, -5, 5]}